Derive conditional density of a random variable? If you have a random variable $X$ that is distributed exponentially with the parameter $\lambda$, how do you condition it on a constant?
For example, $E[X\mid X>y]$. I am confused because $y$ seems completely irrelevant to this distribution. Can we condition on $y$?
 A: Yes, we can condition on the event $X\gt y$. For example, we can ask what is the expected value of $X$, given that $X\gt 1.5$. 
Fix $y\gt 0$. Given that $X\gt y$, we calculate the probability that $X\le w$. This is clearly $0$ if $w\le y$. So let $w\gt y$. We have 
$$\Pr(X\le w\mid X\gt y)=\frac{\Pr(X\le w\cap X\gt y)}{\Pr(X\gt y)}.$$
Compute. The numerator is $(1-e^{-\lambda w})-(1-e^{-\lambda y})$, and the denominator is $e^{-\lambda y}$. So the conditional cdf of $X$ given $X\gt y$ is $1-e^{-\lambda(w-y)}$. We could have obtained this result more simply by referring to the memorylessness property of the exponential.
Thus the conditional density of $X$, given that $Y\gt y$, is $\lambda e^{-\lambda(w-y)}$ (for $w\gt y$). Now we can find the conditional expectation by integrating as usual. 
A: Only one random variable is involved and that is $X$.  There is a tacit universal quantifier, as in "For all values of $y\ge 0$, $\operatorname{E}(X\mid X>y) = \cdots\cdots$".  Thus if $y=6$, you have $$\operatorname{E}(X\mid X>6) = \text{some quantity}$$ and if $y=8$ you have $$\operatorname{E}(X\mid X>8) = \text{some other quantity found by putting 8 where 6 appeared earlier}$$ and in general $$\operatorname{E}(X\mid X>y) = \text{some other quantity found by putting $y$ where 6 or 8 appeared earlier}.$$
It is somewhat conventional, but not universal, to use capital letters near the end of the alphabet to refer to random variables, and lower-case letters as bound variables.  That makes it possible to understand things like $\Pr(X>x)$ and things like the difference between $f_X(3.6)$ and $f_Y(3.6)$, where (capital) $X$ and (capital) $Y$ are random variables.
A: You are confusing expectation conditioned on another random variable, $\mathsf E(X\mid Y)$, and expectation conditioned on an event, $\mathsf E(X\mid X\in A)$, because, yeah, they do look pretty similar.
However, $\mathsf E(X\mid Y)$ is a random variable measured against the sigma algebra generated by $Y$, while $\mathsf E(X\mid X\in A)$ is a value.   In this case:
$$\mathsf E(X\mid X>y) = \frac{\int\limits_{\max(y,0)}^\infty x~\lambda~\mathsf e^{-\lambda x}\operatorname d x}{\int\limits_{\max(y,0)}^\infty \lambda~\mathsf e^{-\lambda x}\operatorname d x} = {\max(y,0)}+\tfrac 1 \lambda$$
Which introduces the memoryless property of the exponential distribution.
$$\mathsf E(X\mid X>y) = y+\mathsf E(X) \qquad \textsf{if }X\sim\mathcal {Exp}(\lambda), y>0$$
